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How can I solve this cubic equation?

$$H^{3} − 3\left[(1 + A\cos(T) )^{2} + \frac{2r \cdot A \sin(T)}{B}\right]H + 2(1 + A \cos(T))^{3} = 0$$

Solution in terms of H.

Edited in order to give more insight to my problem: It was an equation which comes as a part of a derivation in Computational Fluid Dynamics. My motivation is to get H in terms of A, B, r and T. And plot a graph between H and r keeping A and B and T as constants.

Thanks!

A general doubt: If a cubic equation consists of a imaginary root, then is it compulsory that the number of imaginary roots should always be 2?

Later
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bala maverick
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  • I am not sure about the tag and the formatting. Can anyone please help me correct the question? Thanks! @Chandru1: Thanks for the editing! – bala maverick Mar 04 '11 at 14:39
  • Cubic equations have an explicit formula for the roots (like the quadratic formula). Techniques for solving them are discussed here:http://en.wikipedia.org/wiki/Cubic_function There may be easier ways, but you are assured of getting an answer. – Brian Mar 04 '11 at 15:00
  • Are the values $A$, $B$, and $r$ positive? If not, where do they live? – JavaMan Mar 04 '11 at 18:10
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    Where did you get this equation from? What is your motivation? – Abel Mar 04 '11 at 20:50
  • My answer here contains an explicit formula for the solutions to a general monic (leading coefficient 1) cubic that works on most contemporary computing devices/software without any extra worrying about definitions of principal roots. – Isaac Mar 05 '11 at 21:09
  • @DJC: I have Edited the problem to give a better insight.

    @Isaac: Thanks. That was a very nice way of solving cubic equation.. :) Let me see if i can get some intuition from this kind of a solution.

     – bala maverick Mar 07 '11 at 06:18
  • @bala, the fundamental theorem of algebra states that a cubic equation has 3 roots. Any imaginary root must be accompanied by its complex conjugate, so a cubic either has 3 real roots or 1 real root and two complex roots. – rcollyer Mar 08 '11 at 13:05

1 Answers1

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  1. Write it as $H^3 + qH + p = 0$;
  2. use Wolfram Alpha or formulas for roots;
  3. plug in for q and p.
Luke
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TROLLHUNTER
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  • The 2a) is a very Interesting Step! Thank you. Still it would be nice, if i am able to derive it further. because there are 2 imaginary roots and i'll get polynomial of a very high degree for Sin and Cos. Isn't there a better way to simplify the equation? – bala maverick Mar 07 '11 at 06:09
  • A general doubt: If a cubic equation consists of a imaginary root, then is it compulsory that the number of imaginary roots should always be 2? – bala maverick Mar 08 '11 at 04:55
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    @bala maverick: Yes. In general, for a polynomial with real coefficients, any complex roots come in conjugate pairs, so if $x + iy$ is a root for real $x$ and $y$ then $x - iy$ is a root. And since a cubic only has three roots (including multiple roots) there is either a conjugate pair of roots with imaginary parts or no roots with imaginary parts. – Henry Mar 08 '11 at 11:40